Question

Let \(S, T, U\) be three non-void sets and \(f: S \rightarrow T, g: T \rightarrow U\) be so that \(g \circ f: S \rightarrow U\) is surjective. Then
(A) \(g\) and \(f\) are both surjective
(B) \(g\) is surjective, \(f\) may not be so
(C) \(f\) is surjective, g may not be so
(D) \(f\) and \(g\) both may not be surjective

Answer 1

Ans : (B)
Hint : \(\mathrm{g}\) o \(\mathrm{f}: \mathrm{S} \rightarrow \mathrm{U}\) is onto
Let \(\mathrm{z}\) be an arbitrary element of \(\mathrm{U} \because \mathrm{g}\) of \(\mathrm{f}: \mathrm{S} \rightarrow \mathrm{U}\) onto
there exists \(\mathrm{x} \in \mathrm{S}\)
\(\mathrm{g} \circ \mathrm{f}(\mathrm{x})=\mathrm{z} \Rightarrow \mathrm{g}(\mathrm{f}(\mathrm{x}))=\mathrm{z} ; \mathrm{g}(\mathrm{y})=\mathrm{z}\), where \(\mathrm{y}=\mathrm{f}(\mathrm{x}) \in \mathrm{T}\) for all \(\mathrm{z} \in \mathrm{U}\), there exists \(\mathrm{y}=\mathrm{f}(\mathrm{x}) \in \mathrm{T}\) such that \(\mathrm{g}(\mathrm{y})=\mathrm{z}\)
\(\mathrm{g}: \mathrm{T} \rightarrow \mathrm{U}\) onto.

Answer 2

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Answer 3

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